Neither, if the inputs come from voltage sources. The voltage at resistor output would equal to whichever of the two input voltages is higher, minus the voltage drop of the diode. Of course, if the inputs are not fed with a voltage source, the situation is more complex. If the inputs are driven with current sources, then the currents will add up, but currents would add up even without the diodes so they are irrelevant.
How does Logisim handle xor gates without exactly two inputs? A two-input xor gate should have an output which is true if one input is true and the other is false, but it's not completely unambiguous from just that description how that should be generalized to N inputs. The normal approach is to say that the output of an XOR gate of any size should be true if the number of "high" inputs is an odd number; I would consider any other meaning to be non-standard. Nonetheless, it's possible that Logisim is defining "xor(a,b,c)" as being "and(or(a,b,c), nand(a,b,c))" (meaning at least one, but not all, of the inputs is true). Split the 3-input xor into two cascaded two-input xor's and the circuit should behave normally.
It worked fine in the free online version. The only goofy thing was that I had to use these 3-to-1 outputs to get it to allow the output of one gate to go to multiple inputs of the next stage.
Best Answer
Neither, if the inputs come from voltage sources. The voltage at resistor output would equal to whichever of the two input voltages is higher, minus the voltage drop of the diode. Of course, if the inputs are not fed with a voltage source, the situation is more complex. If the inputs are driven with current sources, then the currents will add up, but currents would add up even without the diodes so they are irrelevant.